Boletı́n de la Asociación Matemática Venezolana, Vol. XIX, No. 1 (2012)
71
DIVULGACIÓN MATEMÁTICA
A simple application of the implicit function
theorem∗
Germán Lozada-Cruz
Abstract. In this note we show that the roots of a polynomial are
C ∞ depend of the coefficients. The main tool to show this is the
Implicit Function Theorem.
Resumen. En esta nota se muestra el hecho que las raı́ces de un
polinomio son C ∞ , depende de los coeficientes. La herramienta
principal para mostrar este resultado es el teorema de la función
implı́cita.
1
Motivation
Quadratic equations appear already in antiquity (Babylon, 2000 BC) and
the solution of the general equation, even when it appears in different forms in
the Euclid’s Elements treated by geometric arguments, is obtained algebraically
by the Arab mathematician al-Khowarizmi. He is credited with being the first
to solve the quadratic equation
ax2 + bx + c = 0,
where a, b and c are real numbers with a 6= 0.
Denoting by ∆ = b2 − 4ac, we have the the zeros of (1) are given by
√
−b ± ∆
.
x=
2a
Thus, we have
i) distinct roots if ∆ > 0
ii) equals roots if ∆ = 0 and
∗ Partially
supported by FAPESP, Grant: 09/08435-0, Brazil.
(1)
(2)
72
G. Lozada-Cruz
iii) complex roots if ∆ < 0.
In the case i) we say that the roots of (1) are simple and in the case ii) we
say that the root of (1) is multiple.
A natural question that arises is:
Question 1.1 If we change slightly the coefficients a, b and c in (1), what
happens with the root x given by (1.2)?
First, let’s look at two simple examples:
Example 1.2 Find the roots of the following polynomial p given by
p(x) = x2 + x − 6.
(3)
Using the formula (2) we obtain x = 2 and x = −3. This is the polynomial p
has real and distinct roots.
Example 1.3 Find the roots of the following polynomial p given by
p(x) = (1, 01)x2 + 0, 99x − 6, 01.
Using the formula (2) we obtain x = 1, 99 and x = −2, 97.
polynomial p has real and distinct roots.
(4)
This is the
From (3) and (4) we can see that if we change slightly the coefficients of the
quadratic function then the roots will also change slightly
With the intention of answering this question, we look to the right side of
equation (2) as a function that depends on the coefficients a, b and c. This is
x = x(a, b, c).
(5)
Thus, if we show that x as a function of the coefficients a, b and c is a continuous
function, this answer our question posed above.
More generally consider a polynomial function p : R → R given by
p(x) = a0 + a1 x + · · · + an xn =
n
X
ai xi ,
(6)
i=0
where aj ∈ R, j = 0, 1, · · · , n.
The aim of this short note is to show that the simple real zeros of the equation
(6) depend C ∞ of the parameter λ = (a0 , a1 , · · · , an ) ∈ Rn+1 formed by the
coefficients of p.
A simple application of the implicit function theorem
73
Definition 1.1 We say that c ∈ R is a zero of p if p(c) = 0.
Theorem 1.1 If c is a zero of p then there exists a polinomial function q such
that
p(x) = (x − c)q(x).
(7)
Proof. If c is a zero of p then for all x we have
p(x) = p(x) − p(c) =
n
X
ai (xi − ci ) = (x − c)q(x).
i=0
Definition 1.2 We say that c is a simple zero of p if q(c) 6= 0.
If q(c) = 0 then q(x) = (x − c)r(x), thus p(x) = (x − c)2 r(x), where r is
another polinomial function.
Deriving the equality (7) we have
p0 (x) = q(x) + (x − c)q 0 (x).
(8)
The following result is easy to prove
Theorem 1.2 The number c is simple zero of p if only if p0 (c) 6= 0.
Proof. Follows from (8).
2
C ∞ dependence of the zeros of polynomials with
respect to the coefficients
The main result of this short note is the following result which is the positive
answer to the Question 1.1
Theorem 2.1 If c is a simple zero of p then c is a function of class C ∞ of the
coeficients a0 , a1 , · · · , an of the polinomial p.
Proof. Let us define the following function
f
: R × Rn+1
(x, λ)
→
R
→ f (x, λ)
=
n
P
i=0
where λ = (a0 , · · · , an ).
ai xi
74
G. Lozada-Cruz
Clearly we see that f is C ∞ class in the variables x and λ.
Suppose c0 is a simple zero of polynomial function p0 corresponding to the
particular case λ0 = (a00 , a01 , · · · , a0n ) ∈ Rn+1 then
• f (c0 , λ0 ) =
n
P
i=0
•
a0i (c0 )i = p0 (c0 ) = 0 and
n
∂f
∂ P
(c0 , λ0 ) =
ai xi (c ,λ ) = (p0 )0 (c0 ) 6= 0.
0
0
∂x
∂x i=0
By the implicit function theorem (Theorem 3.1) there are open
neighborhoods U ⊂ R of c0 and V ⊂ Rn+1 of λ0 such that for all λ ∈ V
there exists an unique c ∈ U with f (c, λ) = 0.
Thus, we have a unique map
c:V →U
λ 7→ c(λ)
such that f (c(λ), λ) = 0. This tells us that the polynomial function p
corresponding to the value of the parameter λ = (a0 , · · · , an ) ∈ Rn+1 near
λ0 , also has exactly a simple zero c(λ) ∈ R near to c0 .
Since f ∈ C k for all k ∈ N, the function c(·) ∈ C k for all k ∈ N. This tells
us that the simple zeros of p are C ∞ dependent of the coefficients. Moreover,
the following formula holds
−1
∂f
∂c
∂f
(λ) = −
(c(λ), λ)
·
(c(λ), λ).
∂λ
∂c
∂λ
Remark 2.1 We believe that this simple application of the implicit function
theorem is appropriate for the undergraduate classroom or homework. Also,
using this same theorem we can conclude that the eigenvalues of a matrix A =
(aij )n×n depend continuously of the coefficients aij .
Remark 2.2 This positive result should be contrasted with Abel’s theorem,
which in algebra states that there is not a formula for the zeros of a polynomial
of degree n > 5 in terms of coefficients.
Remark 2.3 For polynomials with more than one variable we have analytical
dependence of the zeros, for more details see [2], p. 362.
A simple application of the implicit function theorem
3
75
Implicit function theorem
For completeness of this note we states here the Implicit function theorem.
More information about the history, theory and applications of this theorem
can be found in [1].
Denoting by
L(Rn ) = {T : Rn → Rn ; T is linear}
Aut(Rn ) = {T ∈ L(Rn ) : T is a bijection}.
Theorem 3.1 (Implicit Theorem) Let W ⊂ Rn × Rp a open set and
f ∈ C k (W, Rn ). Suppose that for some (x0 , y0 ) ∈ W , f (x0 , y0 ) = 0 and
Dx f (x0 , y0 ) ∈ Aut(Rn ). Then there are open neighborhoods U ⊂ Rn of x0
and V ⊂ Rp of y0 such that for all y ∈ V there exists a unique x ∈ U with
f (x, y) = 0.
Thus we have a unique map
ψ:V →U
y 7→ ψ(y)
with ψ(y) = x and f (ψ(y), y) = 0. Moreover, the derivative of ψ in y, Dψ(y) ∈
L(Rp , Rn ) is given by
Dψ(y) = −[Dx f (ψ(y), y)]−1 ◦ Dy f (ψ(y), y),
Proof. See [1].
y ∈ V.
References
[1] Krantz, Steven G.; Parks, Harold R. The Implicit function theorem.
History, theory, and applications. Birkhäuser Boston, Inc., Boston, MA,
2002
[2] Hörmander, Lars, The analysis of linear partial differential operators.
II. Differential operators with constant coefficients. Reprint of the 1983
original. Classics in Mathematics. Springer-Verlag, 2005.
Germán Lozada-Cruz
Departamento de Matemática
IBILCE - Instituto de Biociências, Letras e Ciências Exatas
UNESP- Universidade Estadual Paulista.
e-mail: german@ibilce.unesp.br